Question: Simplify; express your answer in exponential form. Assume $r\neq 0, n\neq 0$. $\dfrac{{(r^{3})^{3}}}{{(rn^{-3})^{-1}}}$
Answer: To start, try working on the numerator and the denominator independently. In the numerator, we have ${r^{3}}$ to the exponent ${3}$ . Now ${3 \times 3 = 9}$ , so ${(r^{3})^{3} = r^{9}}$ In the denominator, we can use the distributive property of exponents. ${(rn^{-3})^{-1} = (r)^{-1}(n^{-3})^{-1}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(r^{3})^{3}}}{{(rn^{-3})^{-1}}} = \dfrac{{r^{9}}}{{r^{-1}n^{3}}}$ Break up the equation by variable and simplify. $\dfrac{{r^{9}}}{{r^{-1}n^{3}}} = \dfrac{{r^{9}}}{{r^{-1}}} \cdot \dfrac{{1}}{{n^{3}}} = r^{{9} - {(-1)}} \cdot n^{- {3}} = r^{10}n^{-3}$.